The temperature at which the rate constants of the given below two gaseous reactions become equal is $\_\_\_\_$ K. (Nearest integer)

$$ \begin{array}{ll} \mathrm{X} \longrightarrow \mathrm{Y}, & \mathrm{k}_1=10^6 e^{\frac{-30000}{\mathrm{~T}}} \\ \mathrm{P} \longrightarrow \mathrm{Q}, & \mathrm{k}_2=10^4 e^{\frac{-24000}{\mathrm{~T}}} \end{array} $$

Given : $\ln 10=2.303$

Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by $20 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If $\mathrm{k}_1$ and $\mathrm{k}_2$ are the rate constants of first and second reaction respectively at 300 K , then $\ln \frac{\mathrm{k}_2}{\mathrm{k}_1}$ will be $\_\_\_\_$ . (nearest integer) $\left[\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$

For the reaction $\mathrm{A} \rightarrow \mathrm{B}$ the following graph was obtained. The time required (in seconds) for the concentration of A to reduce to $2.5 \mathrm{~g} \mathrm{~L}^{-1}$ (if the initial concentration of A was $50 \mathrm{~g} \mathrm{~L}^{-1}$ ) is $\qquad$ . (Nearest integer)

Given : $\log 2=0.3010$

For the reaction A $\to$ products.

The concentration of A at 10 minutes is _________ $\times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$ (nearest integer). The reaction was started with $2.5 \mathrm{~mol} \mathrm{~L}^{-1}$ of A .